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Mice21 [21]
3 years ago
10

Kiara is stacking wood. Kiara stacks sixty-five pieces of wood. There are only seven more pieces of wood for kiara to put on the

stacks. When kiara is done, she looks at the stacks of wood. How many pieces of wood does kiara see in her stack of wood? Show all your mathematical thinking.
Mathematics
2 answers:
Natali [406]3 years ago
8 0
The correct answer is 72
hope i helped you out in some type of way 
Dovator [93]3 years ago
7 0
The answer is 72

You would have to add sixty-five plus seven
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Natasha2012 [34]

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3 years ago
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Write down the set of multiples of 4 between 1 and 25.
jeka57 [31]

Answer:

2,3,4,5,6,7,8,9,10, that's it easy

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Max has two cats. Rex and Roman. If Rex weighs 11 7/18 pounds and Roman weighs 11.329 pounds, which cat weighs the most?​
Vinvika [58]

Answer: Rex weighs more.

Step-by-step explanation:

You have to change both of the numbers into the same form. I did decimal form.

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Roman - 11.329

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<u><em>Hope this helps!</em></u>

8 0
2 years ago
Simplify 12√18-6√20-3√50-8√45
Alex_Xolod [135]

12√18-6√20-3√50-8√45

<em>*Break each radical up into the product of a radical of a perfect square and a second radical*</em>

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<em>*Simplify the radicals*</em>

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<em>*Distribute*</em>

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3 0
3 years ago
The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2
maxonik [38]

Given:

The red figure dilated with a scale factor of s=\dfrac{1}{3} and the center of dilation is at the point (4,2) to get the green figure.

To find:

The coordinates of C' and A.

Solution:

If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

In given problem, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

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Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

4 0
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